Question

A particle is projected vertically upwards from a point O with speed u ms^-1. Two seconds later it is still moving upwards and its speed is 1/3u m s^-1
Find:
a) the value of u.
b) the time from the instant that the particle leaves O to the instant that it returns to O.

Answer 1

Hi,
U have to use linear motion equations here bro..
but before that it would be useful to gather data wt we already have
initial velocity - u m/s
final velocity ( after 2 sec.) - (1/3)*u m/s
gravitational acceleration(g) - 10 m/s^2
time elapsed - 2 seconds

Now the 1st part of the problem asks for the initial velocity (u)
U can use
v = u + at
where,
v = final velocity ( after 2 sec.)
u = initial velocity
a = gravitational acceleration(g)
t = time elapsed

now u will get a equation which will only contain 1 unknown variable(u) and u could simplify it with ur eyes shut.....

Answer 2

For the 2nd part....
it's simply asks the total time that particle spent in the air.

To solve that u must know this
the time taken by vertical projectile to reach it's maximum height is equal to the time that will take by the projectile to reach the ground from the maximum height.

So if we find the time that particle will take to reach it's maximum height we could multiply it by 2 and get the total time of the particle.

Also....
When a vertical projectile is at it;s maximum height it's velocity equals to zero

So now u can use the same equation here again,
v = u + at
where,
v = velocity of the particle at it's maximum height (0 m/s)
u = initial velocity of the particle
a = gravitational acceleration
t = time that particle will take to reach it's maximum height.

for "u" u can use the value u found in the first part....

And finally u could multiply it ( the value for t ) by 2 and get the answer for the second part.

Answer 3

Also,
this is something very important...
Velocity and acceleration r vectors. which mean u have to consider the direction as well

In the first part...
initial velocity is directed vertically upward
but gravitational acceleration is directed vertically downward.
When u r calculation
V = u + at for the 1st part u have to take either up or down direction as positive.
If u take down as positive then initial velocity will be a negative value and g will be positive
V =- u + gt
If u take up as positive then g will be negative while u is positive
V = u - gt

For the second part...
U have to be careful again and do it in the same way as I mentioned above.

I think now u can solve this problem... d u need any more help ?

Hope this will help ya out!

Answer 4

thank u very much

Answer 5

u r welcome bro !